Please find the figure attached below:
Answer:
a.Resistance of bulb=[tex]R_{2}[/tex]=192 ohm
b.power consumed by bulb after motor disconnection=74.609 W
Explanation:
a.Resistance of bulb=[tex]R_{2}[/tex]=?
As we know that [tex]P=\frac{P^{2} }{R}[/tex]
putting R as resistance of bulb i.e. [tex]R=R_{2}[/tex]
[tex]P=\frac{V^{2}}{R_{2} }\\R_{2}=\frac{V^{2}}{P} \\ R_{2}=\frac{(120)^{2} }{75}\\ R_{2}=192ohm[/tex]
b.power consumed by bulb after motor disconnection?
from the figure we see that [tex]R_{1}\ and\ R_{2}[/tex] are in series so
[tex]R_{eq}=R_{1} +R_{2} \\R_{eq}=192+0.5\\R_{eq}=192.5ohm[/tex]
current through their resistance is
[tex]I_{eq}=\frac{V}{R_{eq} }\\ I_{eq}=\frac{120}{192.5 } \\I_{eq}=0.6233A[/tex]
Power consumed by bulb is
[tex]P_{b} =I^{2}R\\\\ P_{b} =(0.6233)^{2}(192)\\\\P_{b} =74.609W[/tex]
